drop table Views;
Create table If Not Exists Views (article_id int, author_id int, viewer_id int, view_date date);
Truncate table Views;
insert into Views (article_id, author_id, viewer_id, view_date) values ('1', '3', '5', '2019-08-01');
insert into Views (article_id, author_id, viewer_id, view_date) values ('3', '4', '5', '2019-08-01');
insert into Views (article_id, author_id, viewer_id, view_date) values ('1', '3', '6', '2019-08-02');
insert into Views (article_id, author_id, viewer_id, view_date) values ('2', '7', '7', '2019-08-01');
insert into Views (article_id, author_id, viewer_id, view_date) values ('2', '7', '6', '2019-08-02');
insert into Views (article_id, author_id, viewer_id, view_date) values ('4', '7', '1', '2019-07-22');
insert into Views (article_id, author_id, viewer_id, view_date) values ('3', '4', '4', '2019-07-21');
insert into Views (article_id, author_id, viewer_id, view_date) values ('3', '4', '4', '2019-07-21');
编写一条 SQL 查询来找出在同一天阅读至少两篇文章的人。结果按照 id 升序排序。
select viewer_id as id
from Views
group by view_date,viewer_id
having count(distinct article_id)>1;